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\(n_{CuO}=a\left(mol\right),n_{Fe_2O_3}=b\left(mol\right)\)
\(m=80a+160b=6\left(g\right)\left(1\right)\)
\(n_{H_2}=\dfrac{2.24}{22.4}=0.1\left(mol\right)\)
\(CuO+H_2\underrightarrow{^{^{t^0}}}Cu+H_2O\)
\(Fe_2O_3+3H_2\underrightarrow{^{^{t^0}}}2Fe+3H_2O\)
\(n_{H_2}=a+3b=0.1\left(mol\right)\left(2\right)\)
\(\left(1\right),\left(2\right):a=0.025,b=0.025\)
\(m_{kl}=0.025\cdot64+0.025\cdot2\cdot56=4.4\left(g\right)\)
\(b.\)
\(m_{hh}=3m_{Fe_2O_3}=6\left(g\right)\)
\(\Rightarrow n_{Fe_2O_3}=\dfrac{2}{160}=0.0125\left(mol\right)\)
\(\Rightarrow n_{CuO}=0.0125\left(mol\right)\)
\(m_{kl}=0.0125\cdot2\cdot56+0.0125\cdot64=2.2\left(g\right)\)
\(n_{CuO}=n_{Cu}=\dfrac{4}{80}=0,05mol\\ n_{H_2}=0,4mol\\ n_{Al}=a;n_{Zn}=b\\ 27a+65b=15,1-0,05\cdot64\\ BTe^-:3a+2b=2\cdot0,4\\ a=0,2;b=0,1\\ m_{ddHCl}=\dfrac{\left(0,6+0,2\right)\cdot36,5}{0,2}=146g\)
a)
$Fe_2O_3 + 3CO \xrightarrow{t^o} 2Fe +3 CO_2$
$Fe + 2HCl \to FeCl_2 + H_2$
$RO + H_2 \xrightarrow{t^o} R + H_2O$
b)
Coi m = 160(gam)$
Suy ra: $n_{Fe_2O_3} = 1(mol)$
Theo PTHH :
$n_{RO} = n_{H_2} = n_{Fe} = 2n_{Fe_2O_3} = 2(mol)$
$M_{RO} = R + 16 = \dfrac{160}{2} = 80 \Rightarrow R = 64(Cu)$
Vậy oxit là CuO
n CO = 6,72/22,4 = 0,3(mol)
n H2 = 2,24/22,4 = 0,1(mol)
B gồm : CO(x mol) ; CO2(y mol)
M B = 18.2 = 36
x + y = 0,3
28x + 44y = 36(x + y)
=> x = y = 0,15
$CO + O_{oxit} \to CO_2$
n O(oxit) = n CO2 = 0,15(mol)
=> m M = 8 - 0,15.16 = 5,6(gam)
n là hóa trị của M
$2M + 2HCl \to 2MCl_n + nH_2$
n M = 2/n . nH2 = 0,2/n (mol)
=> 0,2/n . M = 5,6
=> M = 28n
Với n = 2 thì M = 56(Fe)
n Fe = 5,6/56 = 0,1(mol)
n Fe / n O = 0,1/0,15 = 2/3 . Vậy oxit là Fe2O3
\(\text{Đặt }n_{Al}=x(mol);n_{Fe}=y(mol)\\ \Rightarrow 27x+56y=13,9(1)\\ n_{H_2}=\dfrac{7,84}{22,4}=0,35(mol)\\ a,PTHH:2Al+6HCl\to 2AlCl_3+3H_2(1)\\ Fe+2HCl\to FeCl_2+H_2(2)\\ b,\text{Từ 2 PT: }1,5x+y=0,35(2)\\ (1)(2)\Rightarrow x=0,1(mol);y=0,2(mol)\\ \Rightarrow m_{Al}=0,1.27=2,7(g)\\ m_{Fe}=0,2.56=11,2(g)\)
\(c,n_{HCl(1)}=3n_{Al}=0,3(mol);n_{AlCl_3}=0,1(mol);n_{H_2(1)}=0,15(mol)\\ \Rightarrow m_{dd_{HCl(1)}}=\dfrac{0,3.36,5}{14,6\%}=75(g)\\ \Rightarrow C\%_{AlCl_3}=\dfrac{0,1.133,5}{2,7+75-0,15.2}.100\%=17,25\%\)
\(n_{HCl(2)}=2n_{Fe}=0,4(mol);n_{FeCl_2}=n_{H_2(2)}=n_{Fe}=0,2(mol)\\ \Rightarrow m{dd_{HCl(2)}}=\dfrac{0,4.36,5}{14,6\%}=100(g)\\ \Rightarrow C\%_{FeCl_2}=\dfrac{0,2.127}{11,2+100-0,2.2}.100\%=22,92\%\)
a) 2Al + 6HCl --> 2AlCl3 + 3H2
Fe + 2HCl --> FeCl2 + H2
b) Gọi số mol Al, Fe lần lượt là a,b
=> 27a + 56b = 13,9
\(n_{H_2}=\dfrac{7,84}{22,4}=0,35\left(mol\right)\)
2Al + 6HCl --> 2AlCl3 + 3H2
a----->3a--------->a------->1,5a______(mol)
Fe + 2HCl --> FeCl2 + H2
b------>2b-------->b----->b__________(mol)
=> 1,5a + b = 0,35
=> \(\left\{{}\begin{matrix}a=0,1=>m_{Al}=0,1.27=2,7\left(g\right)\\b=0,2=>m_{Fe}=0,2.56=11,2\left(g\right)\end{matrix}\right.\)
c) nHCl = 3a + 2b = 0,7 (mol)
=> mHCl = 0,7.36,5 = 25,55(g)
=> \(m_{ddHCl}=\dfrac{25,55.100}{14,6}=175\left(g\right)\)
\(m_{dd\left(saupu\right)}=13,9+175-2.0,35=188,2\left(g\right)\)
\(\left\{{}\begin{matrix}m_{AlCl_3}=0,1.133,5=13,35\left(g\right)\\m_{FeCl_2}=0,2.127=25,4\left(g\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}C\%\left(AlCl_3\right)=\dfrac{13,35}{188,2}.100\%=7,1\%\\C\%\left(FeCl_2\right)=\dfrac{25,4}{188,2}.100\%=13,5\%\end{matrix}\right.\)
a, Ta có: 65nZn + 27nAl = 11,9 (1)
PT: \(Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(2Al+6HCl\rightarrow2AlCl_3+3H_2\)
Theo PT: \(n_{H_2}=n_{Zn}+\dfrac{3}{2}n_{Al}=\dfrac{9,916}{24,79}=0,4\left(mol\right)\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\left\{{}\begin{matrix}n_{Zn}=0,1\left(mol\right)\\n_{Al}=0,2\left(mol\right)\end{matrix}\right.\)
⇒ mZn = 0,1.65 = 6,5 (g)
mAl = 0,2.27 = 5,4 (g)
b, Theo PT: nZnCl2 = nZn = 0,1 (mol)
nAlCl3 = nAl = 0,2 (mol)
⇒ m muối = 0,1.136 + 0,2.133,5 = 40,3 (g)
c, Theo PT: nHCl = 2nH2 = 0,8 (mol)
\(\Rightarrow m_{ddHCl}=\dfrac{0,8.36,5}{10\%}=292\left(g\right)\)
n HCl = 360 x 18,25/(100x36,5) = 1,8 mol
H 2 + CuO → t ° Cu + H 2 O
n CuO = x
Theo đề bài
m CuO (dư) + m Cu = m CuO (dư) + m Cu p / u - 3,2
m Cu = m Cu p / u - 3,2 => 64x = 80x - 3,2
=> x= 0,2 mol → m H 2 = 0,4g
Fe + 2HCl → FeCl 2 + H 2
Số mol HCl tác dụng với Fe 3 O 4 , Fe 2 O 3 , FeO là 1,8 - 0,4 = 1,4 mol
Phương trình hóa học của phản ứng:
Fe 3 O 4 + 8HCl → 2 FeCl 3 + FeCl 2 + 4 H 2 O (1)
Fe 2 O 3 + 6HCl → 2 FeCl 3 + 3 H 2 O (2)
FeO + 2HCl → FeCl 2 + H 2 O (3)
Qua các phản ứng (1), (2), (3) ta nhận thấy n H 2 O = 1/2 n HCl = 1,4:2 = 0,7 mol
Áp dụng định luật bảo toàn khối lượng, ta có:
m hỗn hợp + m HCl = m muối + m H 2 O + m H 2
57,6 + 1,8 x 36,5 = m muối + 0,7 x 18 +0,4
m muối = 57,6 + 65,7 - 12,6 - 0,4 = 110,3 (gam)
Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)
a. PTHH:
\(Mg+2HCl--->MgCl_2+H_2\uparrow\left(1\right)\)
\(Cu+HCl--\times-->\)
b. Theo PT(1): \(n_{Mg}=n_{H_2}=0,2\left(mol\right)\)
\(\Rightarrow m_{Mg}=0,2.24=4,8\left(g\right)\)
\(\Rightarrow\%_{m_{Mg}}=\dfrac{4,8}{11,2}.100\%=42,9\%\)
\(\%_{m_{Cu}}=100\%-42,9\%=57,1\%\)
c. Theo PT(1): \(n_{HCl}=2.n_{H_2}=2.0,2=0,4\left(mol\right)\)
PTHH: \(NaOH+HCl--->NaCl+H_2O\left(2\right)\)
Theo PT(2): \(n_{NaOH}=n_{HCl}=0,4\left(mol\right)\)
\(\Rightarrow m_{NaOH}=0,4.40=16\left(g\right)\)
\(A.Mg+H_2SO_4\rightarrow MgSO_4+H_2\\ B.n_{H_2}=\dfrac{1,12}{22,4}=0,05mol\\ Mg+H_2SO_4\rightarrow MgSO_4+H_2\)
0,05 0,05 0,05 0,05
\(\%m_{Mg}=\dfrac{0,05.24}{6,4}\cdot100=18,75\%\\ \%m_{Cu}=100-18,75=81,25\%\\ C.m_{ddH_2SO_4}=\dfrac{0,05.98}{20}\cdot100=24,5g\)
a) Pt : \(Mg+H_2SO_4\rightarrow MgSO_4+H_2\uparrow\)
Theo Pt : \(n_{Mg}=n_{H2SO4}=n_{MgSO4}=n_{H2}=\dfrac{1,12}{22,4}=0,05\left(mol\right)\)
b) \(\%m_{Mg}=\dfrac{0,05.24}{6,4}.100\%=18,75\%\)
\(\%m_{Cu}=100\%-18,75\%=81,25\%\)
c) \(m_{H2SO4}=0,05.98=4,9\left(g\right)\)
\(\Rightarrow m_{ddH2SO4}=\dfrac{4.100\%}{20\%}=20\left(g\right)\)
Chúc bạn học tốt
a) \(n_{H_2}=\frac{7,84}{22,4}=0,35\left(mol\right)\)
\(m_{H_2}=0,35.18=6,3\left(g\right)\)
b) Gọi x là số mol của \(CuO\), y là số mol của \(Fe_2O_3\)
Theo đề bài, ta có: \(80x+160y=20\left(1\right)\)
\(PTHH\left(1\right):CuO+H_2\underrightarrow{t^o}Cu+H_2O\)
\(\left(mol\right)\)________x________________x
\(PTHH\left(2\right):Fe_2O_3+3H_2\underrightarrow{t^o}2Fe+3H_2O\)
\(\left(mol\right)\)______y_________3y___2y_____3y
\(TừPT\left(1\right)và\left(2\right)\Rightarrow x+3y=\frac{7,84}{22,4}=0,35 \left(2\right)\)
\(Từ\left(1\right)và\left(2\right)tacóhpt:\left\{{}\begin{matrix}80x+160y=20\\x+3y=0,35\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0,05\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)
\(m_{hhkl}=m_{Cu}+m_{Fe}=64x+56.2y=14,4\left(g\right)\)
c) \(PTHH\left(3\right):Zn+2HCl\rightarrow ZnCl_2+H_2\)
\(\left(mol\right)\)________0,35___0,7______0,35___0,35
\(m_{Zn}=0,35.65=22,75\left(g\right)\)
\(m_{HCl}=0,7.36,5=25,55\left(g\right)\)
\(m_{ddHCl}=\frac{25,55.100\%}{20\%}=127,75\left(g\right)\)